Statistics for Managers 2
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12
The Chi-Square Test: Analyzing Categorical Data
Learning Objectives
After reading this chapter, you should be able to:
• Describe the conditions that fit chi-square tests.
• Calculate and interpret the goodness of fit test and chi-square test of independence.
• Calculate and interpret the phi coefficient and Cramer’s V.
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CHAPTER 1212.1 Examining Categorical Data
Chapter Outline
12.1 Examining Categorical Data
12.2 The Goodness-of-Fit (1 3 k) Chi-Square Calculating the Test Statistic Interpreting the Test Statistic Understanding the Chi-Square Hypotheses Distinguishing Between Goodness-of-Fit Chi-Square Tests and t-Tests or ANOVAs A 1 3 k (Goodness-of-Fit) Chi-Square Problem With Unequal fe Values A Final 1 3 k Problem
12.3 The Chi-Square and Statistical Power
12.4 The Goodness-of-Fit Test in Excel
12.5 The Chi-Square Test of Independence Setting up the Chi-Square Test of Independence Interpreting the Chi-Square Test of Independence Phi Coefficient and Cramer’s V A 3 3 3 Test of Independence Problem
Chapter Summary
12.1 Examining Categorical Data
The 19th-century British statesman Benjamin Disraeli is credited with saying that there are three kinds of lies: lies, damned lies, and statistics. Clearly, he had to have a place in this book, even if it is in the final chapter. But he belongs here because of another com- ment that is particularly relevant to the topics in this chapter. He observed that what we anticipate seldom occurs and what we least expect generally happens (Oxford, 1980). Disraeli’s expressed skepticism was almost certainly tongue in cheek. Indeed, the work on regression in Chapters 9 and 10 is based on the understanding that outcomes are not unpredictable, but the statement provides an effective segue into the connection between what occurs and what might be expected to occur. That analysis is the focus of this chapter.
Part of the discussion in Chapter 2 was how data differ according to scale, and how the statistics that can be calculated also relate to scale; you learned about different types of data scales and the appropriate types of statistics for each. For example, for nominal scale data, only the mode (Mo) makes sense as a measure of central tendency. Subsequent chap- ters revealed that it is not only descriptive statistics that are specific to the scale of the data. The more involved statistical tests are also data-scale dependent. Recall that the depen- dent variable in a t-test, a z-test, and ANOVA must be data that fit a continuous (interval or ratio) scale. Both variables in the Pearson Correlation must be at least interval scale. These distinctions are very important. Along with whether the hypothesis deals with dif- ference or association and whether the groups are independent, the scale of the data is an important guide to determining the appropriate statistical procedure.
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CHAPTER 12Section 12.2 The Goodness-of-Fit (1 3 k) Chi-Square
Sometimes the best data available are not continuous. There may be no way to verify the normality of the data (perhaps because they are not normal). The groups involved in the analysis may not have equivalent variances. And perhaps the relationships between vari- ables are not linear. When important assumptions about the quality of the data cannot be satisfied, the answer is to move to tests with more relaxed requirements.
Chapter 8 provided the first discussion of nonparametric statistical analysis. Recall that these tests set aside a number of important assumptions about the characteristics of the groups involved in an analysis. Removing some of the requirements associated with the characteristics of the data provides greater analytical flexibility.
Chi-square tests allow for the analysis of data that are exclusively categorical. This distin- guishes them from all the statistical tests discussed in previous chapters. In addition, the use of categorical data, since categorical/nominal data cannot reflect the characteristics of normality, suggests that the chi-square tests are nonparametric tests, also referred to as “distribution-free” tests. No assumptions are needed about how the data are distributed, nor are there requirements regarding their scale. These tests provide flexibility, but they exact a cost as well, which will be discussed as the chapter progresses. The chi-square pro- cedures have many applications in business analysis and decision making and are a main- stay in the manager’s statistical “toolbox.”
The chi-square tests were developed by Karl Pear- son—the “Pearson Correlation” Pearson. Note that the Greek letter for “c” is written x and pronounced with a hard c, so chi is pronounced “kie,” rhyming with “pie.” There are two chi-square procedures discussed in this chapter. Both of them have two names. The first is called the goodness-of-fit chi- square test, or alternatively the 1 3 k (said “one by kay”) chi-square.
12.2 The Goodness-of-Fit (1 3 k) Chi-Square
Perhaps a market research specialist is trying to determine whether several local “talk radio” stations have approximately similar audiences. Among other things, the answer will influence what individual stations can charge for advertising. The market research specialist makes a random selection of people from a local telephone book and calls each number to ask residents if they listen to talk radio. For those answering in the affirmative, the question is which station they prefer.
Both names for this procedure are instructive for what they reveal about the kind of analy- sis involved. “Goodness-of-fit,” however awkward the grammar, indicates that what is at issue is how “good” the data fit an initial hypothesis. That initial hypothesis describes the expected distribution of the data. In the talk radio example, the procedure will be to test whether listeners prefer the major talk stations in about equal proportions; it will provide an analysis of how well the data fit that assumption.
Key Terms: The goodness- of-fit, or 1 3 k chi-square, is a test for significant differences in the categories of a single, nominal scale variable.
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CHAPTER 12Section 12.2 The Goodness-of-Fit (1 3 k) Chi-Square
The 1 3 k chi-square name is a reminder that the procedure involves just a single grouping variable (the “1” in the name), which is divided into some number (k) of categories. Note that the “k” here has the same meaning that it had in ANOVA. It refers to the number of groups in the analysis. In the preceding example:
• The categorical (grouping) variable is the preferred radio station. • The k refers to the number of categories into which the single variable is
divided, which is the number of talk radio stations that will be involved in the analysis.
Take care not to confuse the number of categories with the number of variables involved in the analysis. If there are five talk radio stations in the listening area, there are five cat- egories or dimensions of the single variable, preferred radio station.
Recall that nominal data are also often called categorical, or “count” data, stemming from the fact that the measurement involved with these kinds of data is a matter of asking listen- ers which station they listen to, and then sorting them into the relevant category (preferred radio station) accordingly. This “count” becomes the dependent variable in chi-square tests. The analysis hinges on the frequency with which subjects fall into the individual categories. Note that the issue is not how much more the individual prefers station A to station B, something that would indicate ordinal scale data, and the question is not how much time the individual spends listening, which would provide ratio scale data. The only question for respondents who listen to talk radio is which is the preferred station.
If the question that drives the study is whether listeners prefer the five stations in about equal proportions, that is the hypothesis that will be tested. In that instance, the expecta- tion is that numbers of listeners in each of the categories that represent the different talk radio stations will be reasonably similar. If they are, the resulting chi-square value will not be statistically significant. Statistically significant results emerge in a goodness-of-fit chi- square when there is a substantial discrepancy between what is observed in the data and what is expected based on that initial hypothesis. Exactly how much of a discrepancy there must be is determined by comparing the calculated value of chi-square to a critical value that, like the other t, F, and r test statistics, is determined by the degrees of freedom for the problem and the probability level at which the test is conducted.
Referring back to the radio station problem, the market research specialist needs to either find support for the hypoth- esis that listeners tune in to all five stations in about equal proportions or update the expectation. Ninety-five listeners are asked about their station preferences. Sixty of the respon- dents name one of the five stations in the market area. The other 35 listen to subscription stations on satellite radio that
are not located in the area. Since the interest is in listeners to local stations, those 35 people were excluded from the study. From the remaining 60 listeners, the results are as follows:
Station A 15
Station B 8
Station C 12
Station D 10
Station E 15
Review Question A: What is the scale of the data required by either of the chi- square tests?
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CHAPTER 12
Where
x2 5 the value of chi-square fo 5 the frequency observed; how many individuals occur in the particular
categories fe 5 the frequency expected; how often individuals can be expected to occur in the
particular category according to the initial hypothesis
The chi-square test statistic involves a good deal of repetitive subtracting and squaring. A good way to keep the calculations straight is to complete them in something like Table 12.1. The successive steps in the calculations are represented in the rows beginning with the fo row near the top of the table, and then working down the rows one at a time. Each row in the table represents a calculation step for determining the x2 value.
T
Table 12.1: The 1 3 k chi-square
The survey of the 60 respondents indicates that preferences are as follows:
Station A 15
Station B 8
Station C 12
Station D 10
Station E 15
These results provide a range of 8 listeners for the least frequently mentioned Station B to 15 listeners for the most popular stations, which are Stations A and E, so there are clearly differences in listeners’ preferences. The question the chi-square procedure will answer is whether the differences in “count” across the five stations are just random differences that can be expected because of sampling error, or whether the differences are great enough that they are likely to emerge every time data are collected and the results analyzed. It is that last outcome, of course, that defines statistical significance.
Calculating the Test Statistic
The chi-square test statistic, which is neither intimidating to look at nor difficult to calcu- late, has this form:
Formula 12.1 x2 5 S 1 fo 2 fe 2 2
fe
Section 12.2 The Goodness-of-Fit (1 3 k) Chi-Square
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CHAPTER 12
Following the steps outlined above,
Statistic Station A Station B Station C Station D Station E
fo 15 8 12 10 15
fe 12 12 12 12 12
fo 2 fe 3 24 0 22 3
( fo 2 fe) 2 9 16 0 4 9
( fo 2 fe) 2/fe .75 1.33 0 .33 .75
x2 5 .75 1 1.33 1 0 1 .33 1 .75 5 3.16
The values in the fo row are counts of the number of individuals that occur in each cat- egory of the variable.
• These “frequency observed” values are the number of listeners from the sample of 60 who indicate that they listen to a particular radio station.
• The sum of the fo values across the multiple categories of the variable must always sum to the total sample size, n.
The second row, designated fe, indicates what is expected based on whatever hypothesis or assumption prompted the analysis.
• For the problem above, the hypothesis/assumption is that listeners are attracted to the five stations in approximately equal proportions.
• That expectation is reflected in equal values for each of the different categories.
Later there will be a problem where the expectation is that the categories will not be equal, which must also be reflected in the fe values. If, for example, station A is associated with a major network and has several nationally syndicated shows, perhaps the expectation is that station A is twice as popular as the others. In that case, the fe value for station A would be twice as high as the fe values for each of the other stations.
Here, however, the problem is simpler. The hypothesis is that the stations are equally popular, so determining what to expect is simply a matter of dividing the total number of listeners by the number of categories:
fe 5 n/k
Where
n 5 the total of all subjects in all categories k 5 the number of categories
So for the radio station problem, because n 5 60 and k 5 5, the fe values are determined as follows:
fe 5 60/5 5 12 in each fe category.
Section 12.2 The Goodness-of-Fit (1 3 k) Chi-Square
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CHAPTER 12
Study the formula for the test statistic for a moment, review the order of mathemati- cal operations, and the process for calculating x2 will be straightforward. Recall from what your ninth-grade algebra teacher told you that when there are multiple operations required:
• The first step is to complete whatever needs to be done in the parentheses (“please”).
• Then deal with exponents (“excuse”). • Handle multiplication and division (“my dear”) next. • Then complete any addition and subtraction last (“Aunt Sally”) when they
are not in parentheses.
So in terms of the rows in Table 12.1, the process is to:
1. Fill in the fo values in the first line, determined by the number of people who indicate that they listen to a particular station.
2. For this problem at least, because of the expectation that the five stations are all about equally popular among listeners, divide n by k and enter that value in each of the five fe boxes on the second line.
Now, following order of mathematical operations in columns 1 through 5:
3. On line 3 enter the value that is fo 2 fe for each of the five stations; for each category it is the fo value from the first line minus the fe value from the second line.
4. On line 4 deal with the exponent by squaring each difference between fo and fe. 5. On the next line, divide the result of the previous line ( fo 2 fe)
2 by the fe value from the second line.
6. On the final line, sum the ( fo 2 fe) 2/fe across the five categories to determine the
x2 value.
The result of completing these steps for the data above is that x2 5 3.16.
Interpreting the Test Statistic
The next step is to determine whether a x2 value of 3.16 is statistically significant by com- paring it to the appropriate critical value of x2 from Table 12.2. As with the other tables, the critical value is determined according to the probability level at which the test is con- ducted (p 5 .05 is the default level for this test as well), and the number of degrees of freedom for the problem. The df for the goodness-of-fit chi-square are k 2 1, the number of categories of the variable, minus 1. Here, df 5 4.
Section 12.2 The Goodness-of-Fit (1 3 k) Chi-Square
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CHAPTER 12
Table 12.2: The critical values of chi-square
df p 5 0.05 p 5 0.01 p 5 0.001
1 3.84 6.64 10.83
2 5.99 9.21 13.82
3 7.82 11.35 16.27
4 9.49 13.28 18.47
5 11.07 15.09 20.52
6 12.59 16.81 22.46
7 14.07 18.48 24.32
8 15.51 20.09 26.13
9 16.92 21.67 27.88
10 18.31 23.21 29.59
11 19.68 24.73 31.26
12 21.03 26.22 32.91
13 22.36 27.69 34.53
14 23.69 29.14 36.12
15 25.00 30.58 37.70
16 26.30 32.00 39.25
17 27.59 33.41 40.79
18 28.87 34.81 42.31
19 30.14 36.19 43.82
20 31.41 37.57 45.32
Source: http://home.comcast.net/~sharov/PopEcol/tables/chisq.html Retrieved: 6 July, 2012.
As with the other tests conducted so far, the calculated value of chi-square is statistically significant when it is equal to, or larger than, a critical value. That value is determined by the probability level of the test and the degrees of freedom for the problem. For a x2.05(4) the table indicates that the critical value is 9.49. A critical value from the table greater than the calculated value of chi-square indicates that the fo to fe difference is best explained by differences that could occur by chance in the chi-square distribution. The result is not statistically significant. This result prompts the marketing analyst to fail to reject the null hypothesis.
As an aside, the critical values for chi-square are often appended to two decimals in their tables, just as z values were in their table. With chi-square, stopping at two decimals is not just a matter of crowding more values onto a page as was the case with z. The nominal data upon which chi-square values are based are relatively crude compared to ordinal, or interval or ratio, data, and it makes less sense than with those other data to imply the level of exactness suggested by three decimals.
Section 12.2 The Goodness-of-Fit (1 3 k) Chi-Square
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CHAPTER 12
Understanding the Chi-Square Hypotheses
Like all statistical hypotheses, the null and alternate hypotheses in chi-square problems refer to what occurs in the populations the samples represent.
• In the language of statistics, the null hypothesis for a chi-square problem is that the frequency expected is equal to the frequency observed (Ho: fe 5 fo). The fact that the result in the problem just worked was not statistically sig- nificant indicates that in a population where people listen to five talk radio stations in equal proportions, it is not improbable to draw a sample in which the numbers of listeners who prefer each station range from 8 to 15 out of 60. A sample with fo values of 15, 8, 12, 10, 15 is still consistent with the null hypothesis, and it is one of the outcomes that make up the chi-square distribution.
• The alternate hypothesis is that fo values of 15, 8, 12, 10, 15 is not a sam- ple likely to occur in the chi-square distribution. Stated symbolically it is HA: fe fo; the frequency expected is inconsistent with what was actually observed.
The null hypothesis indicates that any variability between what is observed and what is expected is explained by what will probably occur in the chi-square distribution. In other words, there is insufficient evidence to reject the possibility that what occurred is likely to have occurred by chance. The alternate hypothesis is that there is too much dif- ference between what is observed and what is expected to conclude that the outcome is due to chance.
Distinguishing Between Goodness-of-Fit Chi-Square Tests and t-Tests or ANOVAs
The 1 3 k, or goodness-of-fit, chi-square procedure falls under the general hypothesis of difference category of procedures. In that regard it is similar to the independent samples t-test and to ANOVA. Like those procedures, the value of the chi-square statistic is a mea- sure of difference. The primary difference is the scale of the data in the analysis. In inde- pendent samples t-tests and ANOVA, the t and the F respectively are measures of the difference between the means of the samples involved in the analysis. The independent (grouping) variable is categorical, and the dependent variable is continuous (interval or ratio scale). On the other hand, the chi-square statistic measures the difference between the frequencies of occurrence of a nominal (categorical) variable compared with what is expected. The larger the gap between the expected and observed frequency distribution, the greater the difference between fo and fe. Since the dependent variable is the “count,” or frequency, of occurrence of the categorical variable of interest, it is impossible to calculate means and standard deviations, which makes t-tests and ANOVAs impossible.
A 1 3 k (Goodness-of-Fit) Chi-Square Problem With Unequal fe Values
This first chi-square problem was based on the assumption that listeners preferred the five radio stations in about equal proportions, but equal fe values across all categories of the variable are not always the case. Perhaps a consumer advocate is testing the claim made by the manufacturer of an energy drink called Rush that consumers prefer its product
Section 12.2 The Goodness-of-Fit (1 3 k) Chi-Square
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CHAPTER 12
2-to-1 over the major competitor’s product (Advantage) based on taste alone. If this is accu- rate, a random sample of preferences from consumers of energy drinks should indicate that twice as many prefer Rush over Advantage. Since it is highly unlikely that a random sample will yield exactly those results even if the claim is accurate, the chi-square test can be used to determine whether sample results are close enough to support that claim, or whether results are significantly different from the claim.
The consumer advocate takes a sample of 150 students and finds that 27 of them have used both Rush and Advantage and express a preference for one over the other. The other 123 students prefer either some other energy drink, or use none at all. Their responses are discarded. Of the remaining 27 students, 16 of them prefer Rush and 11 prefer Advantage.
Just as with the first problem, the 11 and 16 numbers represent the fo values, and their sum equals the value of n for the problem. That is the easy part. Because the claim by the manufacturer of Rush is that consumers prefer its drink 2-to-1 over the major competitor, Advantage, the fe values must reflect the 2-to-1 expectation.
Calculating fe Values for Unequal Categories
The total of both frequencies observed and frequencies expected must sum to the total, n. This will always be the case, regardless of the particular hypothesis.
S fo 5 n, and
S fe 5 n
To calculate the fe values when the numbers in multiple cat- egories are not the same will involve vindicating that ninth- grade math teacher who said that someday algebra would be helpful. To determine the fe values,
1. Let x equal fe for the number who prefer the Advantage energy drink 2. Since the expectation in this example is that twice as many consumers will
prefer Rush over Advantage, let 2x be the fe for those who prefer the Rush energy drink
3. Because the fe categories must sum to the total then x 1 2x 5 n 4. Since, n 5 27, the expression can be changed as follows: x 1 2x 5 27
a. If x 1 2x 5 27, if follows that 3x 5 27 b. If 3x 5 27, then x 5 27/3, which makes x equal to 9
As a result of these calculations, the fe value for the Rush consumers is 18 (2x 5 2 3 9 5 18), and the fe value for Advantage consumers is 9 (x 5 9). With those values in hand, the claim that Rush is preferred twice as often as Advantage on the basis of taste can be tested with the 1 3 k chi-square. The solution is Table 12.3.
Review Question B: What is the null hypothesis in a chi- square problem?
Section 12.2 The Goodness-of-Fit (1 3 k) Chi-Square
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CHAPTER 12
Table 12.3: A 1 3 k chi-square for unequal fe values: Is Rush twice as popular as Advantage?
Advantage Rush
fo 11 16
fe 9 18
fo 2 fe 2 22
( fo 2 fe) 2 4 4
( fo 2 fe) 2/fe .44 .22
x2 5 .44 1.22 5 .66
x2 5 .66
x2.05(1) 5 3.84. Accept Ho.
Interpreting the Results
Since the calculated value of chi-square is lower than the critical value from the table for p 5 .05 and 1 degree of freedom, the decision is to fail to reject. That part is straight- forward enough, but in this problem where the claim is that Rush is twice as popular as Advantage, what does failing to reject mean? The key is the null hypothesis, which always reflects the expectation upon which the test is based. Because this problem was set up with an fe value for one outcome that is two times the value of the other, failing to reject the null hypothesis means that there is not enough evidence to reject the claim that Rush is twice as popular as Advantage. To say it another way, although the data do not reflect exactly a 2-to-1 preference for Rush (16 is not 2 3 11), the departure from that claim is not sufficient to allow the consumer advocate to reject it.
Note that the makers of Rush maintain that their product is twice as popular as Advantage based on taste. Whether it is entirely taste or not probably cannot be verified. Perhaps it is marketing ability that prompts the students to prefer Rush, or maybe the costs of the two products differ, or maybe one comes in a more convenient size than the other. The way the data were collected made the consumer’s stated preference the issue, without questions about the reasons for the preference. For whatever reason, students prefer the one product to the other by a great enough margin that it could be 2-to-1.
A Final 1 3 k Problem
To solidify the grasp of the goodness-of-fit procedure, here is one more problem. In this instance a consulting company is retained by a satellite provider to determine whether, in a particular region of the country, satellite TV is three times more popular than free TV, and whether cable TV is twice as popular as free TV. The consulting company is retained to check the veracity of that claim. A random sample of 93 viewers in the region is exam- ined and found to rely on the following for television service:
Section 12.2 The Goodness-of-Fit (1 3 k) Chi-Square
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CHAPTER 12
Satellite 65
Cable 16
Free 12 The same approach used in the last problem to determine the fe values produces the following:
3x (satellite) 1 2x (cable) 1 x (free TV) 5 93
6x 5 93
x 5 15.5
That x value makes the fe for
free TV 5 15.5, for cable TV 5 2 3 15.5 5 31, and for satellite TV 5 3 3 15.5 5 46.5.
Note that the fe values sum to 93, as they must. Do not be distracted by the fact that the fe values are not whole numbers. Although asking which type of TV service people are using will make the fo values whole, the fe numbers indicating what people are expected to use can take on any value. The calculations for this problem are in Table 12.4.
Table 12.4: Another 1 3 k chi-square problem
Satellite Cable Free TV
fo 65 16 12
fe 46.5 31 15.5
fo 2 fe 18.5 215 23.5
( f o 2 fe) 2 342.25 225 12.25
( f o 2 fe) 2/fe 7.36 7.26 .79
S x2 5 7.36 1 7.26 1 .79 5 15.41
With a calculated x2 5 15.41 and the table value for x2 of 5.99 when testing at p 5 .05 with 2 degrees of freedom, the results indicate that the null hypothesis should be rejected. The way these 93 people are distributed does not fit a chi-square dis- tribution where cable is twice as popular and satellite is three times as popular as free television.
Review Question C: How is the scale of the data involved in an analysis related to the power of the statisti- cal procedure?
Section 12.2 The Goodness-of-Fit (1 3 k) Chi-Square
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CHAPTER 12Section 12.3 The Chi-Square and Statistical Power
The difficulty with a problem such as this one is that the initial hypothesis is actually two hypotheses: Satellite is three times as popular as free television, and cable is twice as pop- ular as free television. When the results indicate rejecting the null hypothesis, it could be because the expected ratio of satellite versus free TV customers is not supported, because the ratio of cable versus free TV customers is not supported, or because neither hypothesis is supported. Recall that this was the case with a significant F in ANOVA as well. It did not provide clear evidence for which two specific groups are significantly different. A more definitive chi-square test would require that the consultant gather new data and check those hypotheses individually.
12.3 The Chi-Square and Statistical Power
As noted at the beginning of the chapter, distribution-free tests like chi-square proce-dures provide great flexibility. They provide no restrictions regarding the scale of the data, there are no normality assumptions to contend with, and these procedures work quite well with small samples. In the statistical version of the “there’s no such thing as a free lunch,” expression, however, chi-square and most nonparametric procedures have a drawback.
Note that even though there were differences between what was observed and what was expected in the first two problems in this chapter, neither chi-square value was statistically significant. The chi-square procedures are not very sensitive to minor variations between what is seen and what is expected. Indeed, the differences must be fairly substantial to produce a significant chi-square value. Recall that power in statistical testing is the ability of a procedure to detect statistical significance. Compared to something like ANOVA, the chi-square procedures are not particularly powerful.
Much of the lack of power comes down to the scale of the data that are involved. Nominal data provide no information about what is measured except the category of the variable to which the individual belongs. For each of the energy drink consumers who prefer Rush over Advantage, all that is revealed is which energy drink is preferred. Nothing in the data indicates how much more one drink is preferred over the other. There is no ranking of preference on a scale of 1 to 5. All that is known is that, presented a choice, the consumer chose Rush.
Recall that all statistical tests are based on probabilities, and that because the outcome is therefore never a certainty, there is the constant possibility of a Type I or Type II decision error. Because the chi-square procedures are insensitive to minor variations between fo and fe, chi-square analyses are more inclined toward Type II (beta) decision errors than tradi- tional parametric tests of significant differences. The risk is that an analysis that suggests that results are not statistically significant might be set aside if new data were gathered and the analysis run a second time.
On the bright side, Type I (alpha) decision errors are relatively uncommon. It is not likely that upon finding a result statistically significant, further testing with new data would suggest otherwise; a decision to reject the null hypothesis is unlikely to be overturned by a second analysis.
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CHAPTER 12Section 12.4 The Goodness-of-Fit Test in Excel
12.4 The Goodness-of-Fit Test in Excel
The procedures in the Excel Data Analysis package do not include the chi-square tests. However, because the test statistic involves a good deal of repetitive subtracting, squaring, dividing, and so on, it is not difficult to set up an Excel spreadsheet to accommo- date a 1 3 k problem. This can be done by organizing a spreadsheet to complete the same calculations used in Tables 12.1, 12.3, and 12.4. To illustrate, an organic vegetable grower claims that in spite of the higher price, shoppers will select organically grown spinach as often as spinach grown with the help of pesticides and chemical fertilizers. The first 30 people who buy spinach on a particular day at a grocery store are examined for whether they bought the organically grown vegetable. Results are as follows:
Organically grown: 10 Conventionally grown: 20
To test the grower’s claim,
• Enter the labels organic in cell B1 and conventional in C1. • Enter the labels fo in cell A2, fe in A3, fo 2 fe in A4, ( fo 2 fe) sqd in A5, 4fe in
A6, and sum in A7. • Enter the values 10 and 20 in cells B2 and C2 respectively.
Since the claim is that organic spinach will sell as frequently as conventionally grown spinach, the fe values are simply n 4 2 5 15.
• Enter 15 in both B3 and C3. • In cell B4 enter the formula 5B2-B3 and then press Enter. • With the cursor on B4 hold the shift key down and move the cursor to C4 and
then enter the command to fill right which will be near the far right of the menu ribbon so that the procedure in B4 is repeated in C4.
• In cell B5 enter the command 5B4^2 to square the value in B4. • Repeat the B5 procedure in C5 using the fill right command as above. • In cell B6 enter the command 5B5/B3. • Repeat the B6 procedure in C6 using the fill right command. • In cell B7 enter the command 5sum(B6:C6).
The last command in the above sequence produces the chi-square value for this problem, x2 5 3.33. The critical value from the table for 1 degree of freedom and p 5 .05 is x2 5 3.84. The results are not significantly different from the organic grower’s claim; the organically grown spinach may be just as popular as the traditionally produced spinach, in spite of its higher price. Figure 12.1 is a screenshot of the spreadsheet for this problem.
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CHAPTER 12Section 12.5 The Chi-Square Test of Independence
Figure 12.1: Chi-square goodness-of-fit using Excel
12.5 The Chi-Square Test of Independence
The goodness-of-fit or 1 3 k chi-square procedure accommodates just one categorical (grouping) variable. In that regard, it is similar to a one-way ANOVA, which likewise involves just a single variable, although it is an interval or ratio scale rather than a nomi- nal scale variable with the data divided into any number of categories or groups. Although basing an analysis on a single variable keeps the arithmetic simple, it consigns to error any variance that is not explained by that single variable. On the other hand, when multiple independent variables are included, as is the case in factorial ANOVA, there is less residual variance and a smaller error term. In addition, besides each variable contributing to the result, sometimes multiple variables act together, a phenomenon called a “statistical inter- action” in factorial ANOVA. A similar thing can happen with chi-square procedures. Some- times a single variable is an inadequate explanation of an outcome, and in those circum- stances a second variable will act in concert with the first.
The factorial ANOVA has an approximate equiva- lent in one of the chi-square procedures. It is called the chi-square test of independence, or the r 3 k chi-square. The names for this procedure are just as informative as were those for the one-variable chi-square test. The “test of independence” alludes to the fact that what is tested is whether the two vari- ables included in the analysis operate independently.
Key Terms: The chi-square test of independence, or r 3 k chi-square is a test of the inde- pendence of two nominal scale variables.
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CHAPTER 12Section 12.5 The Chi-Square Test of Independence
In the language of ANOVA, the variables are analyzed to determine whether they interact. When the chi-square value is statistically significant, it indicates that the two variables do not operate independently, a result that will lead to an ancillary analysis to determine the level of their relationship.
As was the case in factorial ANOVA, both of the interacting variables in the r 3 k chi- square are categorical. The r 3 k designation refers to the way the analysis is set up. The data are organized so that the levels of one variable are indicated in the rows (r) of a table, and the other variable is represented in the columns, with each column representing a separate category (k) of that variable. Although the calculations will fit the same table that was used for the goodness-of-fit (1 3 k) problems, the frequency expected ( fe) values are calculated differently.
Setting up the Chi-Square Test of Independence
The rows and columns explanation just above pro- vides the organization for a contingency table that is used in this chi-square test of independence. The way it is used can be illustrated with a problem. The human resources department for a fast food chain is considering offering an early retirement package to some of the more senior managers in an effort to reduce payroll. The department wishes to be able to predict whether a severance package with a $15,000 bonus offered to early retirees will affect retirement plans. Because of the potential costs associated with offering the bonus to dozens of senior managers, the human resources people need to have some understanding of the impact that the bonus will have on employees’ decisions to retire. Among the managers, 30 senior managers are identified and randomly divided into two groups of 15 each.
• The 15 managers in the first group are asked to complete a questionnaire, to be submitted anonymously, which includes a question about whether the respondent anticipates retiring within the next three years. Among this group, two managers indicate that they intend to retire within the specified period.
• Those in the second group of 15 managers are asked whether, if a $15,000 bonus were offered to those who retire in the next three years, they would retire in that period. Of the 15, seven managers indicate that they would retire within the next three years if the bonus were offered.
Note that there are two potentially related variables involved. One is whether managers intend to retire in the coming three years. The other is whether they would retire in that time frame if a bonus were offered. These two variables can be represented in a table that looks much like what was used earlier to set up a two-way ANOVA. In addition to helping one to visualize the problem, when used in the chi-square test of independence the table helps with the task of deriving the fe values. But before worrying about the calculations, note the contingency table below. It is organized with the categories of one variable in the rows of the table and the categories of the other variable in the table columns:
Key Terms: The contingency table organizes data into rows for the categories of one variable, and columns for the catego- ries of the other in an r 3 k chi-square.
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CHAPTER 12Section 12.5 The Chi-Square Test of Independence
Retirement Yes Retirement No No Bonus
Bonus
Organizing the Contingency Table
This same table appears in Table 12.5 with the results of the survey filled in. There are also totals for each row, each column, and a value for all subjects together, n. This particular contingency table is a 2 3 2. Although the chi-square test of independence is limited to two variables, those two variables can each have any number of categories. There could have been five levels of the bonus, for example offered to different groups of potential retirees: no bonus, a $5,000 bonus, a $10,000 bonus, a $15,000 bonus, and a $20,000 bonus. There might have been more than two categories of the retirement decision as well: retire in the next year, retire in two to three years, retire in four to five years, and so on. With the 2 3 2 problem, there are just four cells labeled “a” through “d.” Note that the value in cell “a” is the number that represents the combination of the no-bonus group and the number in that group who indicated that they would retire. The number in cell “d” is the combination of the bonus and the number in that group who opted not to retire, and so on.
Table 12.5: The chi-square test of independence for retirement decisions and a severance bonus
A. The Contingency Table
Will Retire Won’t Retire Row Totals
No Bonus a 2 b 13 15
Bonus c 7 d 8 15
Column Totals 9 21 n 5 30
B. Completing the Analysis
Statistic a b c d
fo 2 13 7 8
fe 4.5 10.5 4.5 10.5
fo 2 fe 22.5 2.5 2.5 22.5
( f o 2 fe) 2 6.25 6.25 6.25 6.25
( f o 2 fe) 2/fe 1.39 .60 1.39 .60
x2 5 1.39 1 .60 1 1.39 1 .60 5 3.98
The two sample groups each have values that sum to 15, a value reflected in the row totals on the right. The sum of the two rows must total n. Although the 2 columns together must also sum to 30, the individual columns will not necessarily each be 15 since the number opting for and the number opting against retirement from the bonus and no-bonus groups are not equal.
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CHAPTER 12Section 12.5 The Chi-Square Test of Independence
The fo and fe Values in the Chi-square Test of Independence
The values in each of the four cells are the fo values that will be used to calculate the chi- square value. Note that in Part B of Table 12.5, the fo values are listed just as they were listed in the goodness-of-fit problems done earlier.
The difference between the way the chi-square values are calculated in goodness-of-fit and test of independence problems is in the way the fe values are determined. For each of the “a” through “d” cell values, fe is calculated as the row total in which the particular cell is included times the column total in which the cell is part, divided by n. Symbolically, for each cell fe 5 (row ttl 3 col ttl)/n. This makes the fe values in the retirement problem as follows:
• Cell a: (15 3 9)/30 5 4.5 • Cell b: (15 3 21)/30 5 10.5 • Cell c: (9 3 15)/30 5 4.5 • Cell d: (15 3 21)/30 5 10.5
Once the fe values are determined, the rest of the calculations for a chi-square value are the same as they were for a goodness-of-fit test and are completed in Part B of Table 12.5. Begin by subtracting fe from fo, square the difference, and so on. The critical value of chi- square comes from the same table used for goodness-of-fit problems. The degrees of free- dom for the test are determined by taking the rows minus 1, times the number of columns minus one (df 5 (r 2 1) 3 (k 2 1)). For this problem, df 5 (2 2 1) 3 (2 2 1) 5 1.
With the calculated value x2 5 3.98, and the table value x2.05(1) 5 3.84, the result is statistically significant.
In the context of the r 3 k procedure, what does a significant outcome mean? Chi-square results are statistically significant when the fe values diverge enough from the fo values that the difference between the two is unlikely to have occurred by chance. The implica- tion is that the factor that creates the fo versus fe difference in r 3 k problems is the rela- tionship between the two variables. If the retirement decision and the availability of the retirement bonus variables are unrelated, which is to say that they operate independently, there will be no significant result. The fo 5 fe null hypothesis for an r 3 k problem has the same meaning as the null hypothesis in a Pearson Correlation problem. It means that there is no relationship between the two variables. The difference is that a Pearson Cor- relation cannot be calculated between two nominal variables.
The Yates Correction to 2 3 2 Problems
Earlier we said that Type I decision errors are relatively uncommon with chi-square proce- dures. While that is generally true, the 2 3 2 problem, where each variable has two levels like the example here, may be an exception. With those problems there can be a tendency to incorrectly find statistical significance when one or more of the fe values in the problem fall below 5.0. In what is now called the “Yates correction,” Yates suggested curbing this tendency by subtracting .5 from all fo 2 fe cell differences in any 2 3 2 problem if at least one fe value is less than 5.0. The reduced fo 2 fe difference makes a significant chi-square value less likely, of course, and so reduces Type I error probability.
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CHAPTER 12Section 12.5 The Chi-Square Test of Independence
There is not a consensus about the use of the Yates correction, and some statisticians argue that the .5 reduction is in fact an overcorrection that makes the procedure unnecessarily conservative. The decision in this book has been to not incorporate the correction. The issue is raised so that the reader will know what it is and why some analysts recommend making the correction. For more information, Howell (1992) provides a helpful discussion of the Yates correction.
Interpreting the Chi-square Test of Independence
The issue in both the chi-square goodness-of-fit (1 3 k) and test of independence (r 3 k) is whether what is observed is consistent with an expected outcome. In the case of the test of independence, a significant result (rejecting the null hypothesis) indicates that the two variables are not functioning independently; they are correlated. In our example, rejecting Ho indicates that the intention to retire and the availability of the cash bonus are related. In a reference back to the difference between correlation and causation in Chapter 8, it is not clear that the bonus causes managers to make a retirement decision, but for whatever reason, there are significantly more intents to retire when the bonus is part of the equation.
At this point the focus turns to the nature of that relationship. Since it is clear from com- paring the calculated chi-square value to the table value that there is a correlation, the question now is of the strength of the relationship between the two variables.
Phi Coefficient and Cramer’s V
To this point the analysis was similar to ANOVA or t-tests and fell under the general umbrella of the hypothesis of difference. Having determined a significant difference between fo and fe, the focus now shifts to a hypothesis of association issue.
The correlation procedure for interval/ratio variables that meet normality requirements was Pearson’s r. For a correlation of ordinal scale data, or for interval/ratio data that fail to satisfy normality requirements, Spearman’s rho was the answer. The need here is for a correlation procedure based on nominal data, and there are several from which to choose. Pearson, who developed chi-square, also developed a correlation procedure for nominal variables called coefficient of contingency, C. Because it produces quite a con- servative correlation value, it is not as widely used as some of the alternatives. The upper bound for most correlation procedures is 1.0. The coefficient of contingency can- not reach that value.
Two of the other correlation procedures for nomi- nal variables are phi coefficient, f (f is the Greek equivalent of f ) , and Cramer’s V, which are both explained here. Contingency coefficient, phi coeffi- cient and Cramer’s V are all based directly on the chi-square value, which makes them easy to calcu- late once the chi-square value has been determined.
Key Terms: Phi coefficient and Cramer’s V are both cor- relation procedures for nominal data used after a significant r 3 k chi-square result.
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CHAPTER 12Section 12.5 The Chi-Square Test of Independence
Formula 12.3 V 5 "1f2/fewer of rows or columns 2 1 2
Where
x2 5 the value calculated in the r 3 k procedure n 5 the total number of subjects
For the decision-to-retire and the cash bonus problem, x2 5 3.98, and n 5 30. Therefore to solve for f:
f 5 "1x2/n 2 5 "13.98/30 2 5 "1.33
f 5 .36
The retirement/cash bonus relationship is f 5 .36. Although f cannot have a negative value (the ways x2 is calculated and the square root function in the f formula do away with that possibility), it is interpreted like any other correlation statistic. A 0 correlation indicates no relationship; a 1.0 correlation indicates a perfect correlation. The correlation here of f 5 .36 is “modest,” or perhaps “low,” by correlation standards.
When either of the two variables in the analysis has two levels, V 5 f. This is because the formula for Cramer’s V is,
Furthermore, if the chi-square value is statistically significant, the correlation coefficient will be significant at the same level; there are no separate significance tests necessary for C, V, or f. If the chi-square value is not statistically significant (if the decision in the initial chi-square analysis is to fail to reject), there is no point in calculating a correlation value since failing to reject Ho: fo 5 fe indicates that the variables are independent.
In a statistically significant 2 3 2, 2 3 3, or 3 3 2 chi-square problem, phi coefficient will be the appropriate follow-up correlation procedure. The formula is the following:
Formula 12.2 f 5 "1x2/n 2
Where
f2 5 the square of the phi coefficient rows or columns 5 the number of levels of the two variables
If there are just 2 levels of either variable, the divisor is 1 and ".362 5 .36, V 5 f. If the fewest number of rows or columns is three, this changes, of course, and the correct correla- tion value to calculate is Cramer’s V.
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CHAPTER 12Section 12.5 The Chi-Square Test of Independence
A 3 3 3 Test of Independence Problem
A property management company in a large city manages the landscaping and rent col- lection at several apartment complexes. Some of the complexes are quite large, with more than 50 units; some are very small, with fewer than 10 units; and the others are classi- fied as medium-sized. Collecting and crediting rent payments each month is very time- consuming. A bookkeeper at the management company guesses that in the smaller complexes there is a more intimate relationship between manager and tenant than in the larger complexes, and rent difficulties are correspondingly lower as a result. To test that assumption, the bookkeeper examines data from 100 apartments located in each of small, medium, and large complexes and determines the number of rent payments that are on time, that are within one week late, and that are more than one week late. The data are below:
Rent Submission
On-time Within 1 week .1 week late Row totals
Small a 65 b 30 c 5 100
Medium d 55 e 35 f 10 100
Large g 45 h 25 i 30 100
Column totals 165 90 45 300
The first question is whether the two variables of apartment complex size and rent late- ness are independent. If the chi-square value is statistically significant, the decision will be that they are not independent, and that will prompt a second question about the strength of their relationship. The calculations for this chi-square test of independence are in Table 12.6.
Table 12.6: A chi-square test of independence for the size of the apartment complex and the lateness of the rent
a b c d e f g h i
fo 65 30 5 55 35 10 45 25 30
fe 55 30 15 55 30 15 55 30 15
fo 2 fe 10 0 210 0 5 25 210 25 15
( fo 2 fe) 2 100 0 100 0 25 25 100 25 225
( fo 2 fe) 2/fe 1.82 0 6.67 0 .83 1.67 1.82 .83 15
S 28.64
• The calculated x2 5 28.64. Since this is a 3 3 3 problem, df 5 (3 2 1) 3 (3 2 1) 5 4.
• The critical value of chi-square x2.05(4) 5 9.49.
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CHAPTER 12Section 12.5 The Chi-Square Test of Independence
The result is statistically significant. How promptly renters pay their monthly rent is related to the size of the complex in which they live.
Determining the strength of the correlation calls for Cramer’s V since both variables have more than two levels.
V 5"1f2/fewer of rows or columns 2 1 2
But since V requires the calculation first of phi coefficient, the answer begins there.
f 5 "1x2/n 2
5 "128.64/300 2
5 .31
With a value for phi, V can be calculated.
V 5"1f2/fewer of rows or columns 2 1 2
5 "1 .312/2 2
5 ".05
5 .22
The relationship between the size of the rental complex and how promptly rent is paid is V 5 .22. The correlation is not particularly robust, but it is statistically significant, since the value of chi-square upon which it is based is significant. Based on the analysis, those at the property management company
are in a position to alter procedures in some way that responds to the relationship between rent payment and complex size. Perhaps it will make a difference if rent collections can be made online. Maybe an effort to improve the social relationship between the apartment manager and the tenants, particularly in large apartment complexes, will prompt rent payments to be made in a more timely fashion.
Review Question D: What does phi coef- ficient measure?
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CHAPTER 12Chapter Summary
Chapter Summary
The chi-square tests assume that Disraeli was unnecessarily skeptical. In fact, an informed expectation of an outcome should provide a fairly good indicator of what will actually occur, which is the understanding upon which both chi-square tests are based. The chi- square tests answer many of the same questions that earlier tests in this book answered. The difference is that the chi-square tests are based on nominal data. Pearson developed tests for data which indicate nothing more than the count of the number that occur in a particular category. Consequently, the analysis is based on differences between the fre- quency observed ( fo) and the frequency expected ( fe) (Objective 1).
The goodness-of-fit, or 1 3 k procedure analyzes whether the proportions occurring in the multiple categories of a single variable are consistent with what is expected based on an initial hypothesis. The chi-square test of independence, also called the r 3 k procedure, straddles the boundary between tests of the hypothesis of difference and those related to the hypothesis of association. The initial analysis establishes whether two variables func- tion independently. Like the goodness-of-fit test, this part of the analysis is based on the magnitude of the fo 2 fe difference. A significant value of chi-square means rejecting the probability that the variables are independent (Objective 2). At that point the question is about the strength of the relationship between the two variables. That correlation can be gauged by one of several correlation procedures designed for nominal data. Those cov- ered in this chapter include the phi coefficient and Cramer’s V (Objective 3).
It should be noted here that this book represents only a brief introduction to the analysis procedures that can be useful for managers. The list of statistical procedures covered in 12 chapters is far from exhaustive, but it is a valuable beginning. The different tests explained in Chapters 1212 are representative of those that are appropriate to many kinds of busi- ness analysis. Figure 12.2 is a flowchart-like guide to which test will answer the manager’s question. It is provided here as a summary and overview of the preceding chapters. As the decision tree is followed from the top down, note the issues are:
• Is the question about differences or associations? • Are the data involved nominal (categorical), ordinal, or interval/ratio? • How many groups are involved? • Are the groups independent?
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CHAPTER 12Chapter Summary
Figure 12.2: Finding the appropriate test
Answering each question above in turn guides one to the test tailored to the particular problem. If the question is about differences between groups (1), the measures involved are interval or ratio scale (2), there are 4 groups involved (3) that are independent (4), one of the ANOVA tests will answer the question.
No statistics book can provide comprehensive coverage of every statistical test, and early in the development of this book a decision was made to present tests neither for sig- nificant differences nor association for ordinal data. They are presented in Figure 12.2 to round it out, but they are not found elsewhere in the book. Should the reader wish to pur- sue Mann-Whitney, Kruskal-Wallis, Wilcoxon, or Friedman’s ANOVA tests, Tanner (2011) is a useful source.
No. Groups The testIndependent?
Independent
Related
Questions about Differences Questions about Associations
Chi-square goodness of fit Phi Coefficient
Spearman’s rho
Pearson Correlation
Point-biserial Correlation
Multiple Correlation
Semi-partial Correlation
Chi-square test of independence
Mann-Whitney U
Wilcoxon T
Kruskal-Wallis H
Friedman’s ANOVA
Friedman’s ANOVA
Independent t
Before/After t
Analysis of Variance
1
2
2
2+
Data Scale
Nominal data
Ordinal data
Interval/ Ratio data
Independent
Related
Independent
Related
2
2+ Independent
Related
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CHAPTER 12Chapter Formulas
When the procedures encountered in this book are used by managers with an under- standing of the procedures’ purpose and an appreciation for their requirements, they offer the promise that the related decisions will be more reasonable and better informed, and can equip managers with the tools they need to be most effective. Finally, if you wish to broaden your horizons in the future, virtually all of the more advanced procedures that are beyond the scope of this book are based on the concepts represented here. Your authors wish you the best of luck. The first author can be reached with comments and questions at [email protected].
Answers to Review Questions
A. The chi-square procedures require data of only nominal scale. B. The null hypothesis is that the frequency observed equals the frequency expected,
fo 5 fe, meaning that there is not enough evidence to reject the possibility that what emerged in the analysis was consistent with the prediction.
C. Nominal data provide very little measurement information, and the chi-square procedures are a case in point. The analyses are based on nothing more than the frequency with which data occur in the various categories. These data yield noth- ing about how much of a measured quality is present, for example. Because none of the data nuances present with ordinal and interval/ratio data are gauged, differences must be substantial to prompt rejecting the null hypothesis. These are not particularly powerful procedures.
D. Phi coefficient measures the strength of the relationship between two nominal variables when at least one of them has only two categories.
Chapter Formulas
Formula 12.1 x2 5 S 1 fo 2 fe 2 2
fe
is the formula for the chi-square test statistic. The same formula is used for both “goodness of fit” test and for the r 3 k chi-square test of independence.
Formula 12.2 f 5"1x2/n 2
Phi coefficient is the measure of the correlation for two nominal variables when the chi- square test of independence indicates a significant result, and when one of the variables involved has two or fewer categories.
Formula 12.3 V"f2/ 1smaller of rows or columns 2 2 1
When the test of independence is significant and both variables have at least three catego- ries, Cramer’s V is calculated rather than phi coefficient. V requires phi, however, which must be calculated first.
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CHAPTER 12Management Application Exercises
Management Application Exercises
Unless otherwise stated, use p 5 .05 in all your answers.
1. Three new movies, each with the potential to be a blockbuster, are released on the same day. Reporters from the local television station are interested to see whether one appears to have caught the public attention more than the others. The reporter goes to the local multiplex and asks those waiting to buy tickets which movie they intend to see. On the basis of results from 52 people, are there significant differ- ences in movie preferences? The data are as follows:
Fantasy Haven: 22 Night of Terror: 18 Fists of Glory: 12
2. Data from behavioral psychology indicate that administering a tangible reward to subjects will prompt response levels twice as frequent as from subjects who receive a nontangible reward. To test this notion in a business context, two sales seminars are compared. In one seminar, sales representatives are tossed a piece of candy every time they ask a relevant question or provide an insightful comment. In the other seminar, only verbal reinforcement is provided. At the end of the seminars, data are as follows:
verbal reinforcement seminar—17 questions/comments tangible reward seminar—27 questions/comments
a. What is the fe value for each group? b. Are the results consistent with the expectation?
3. A Department of Labor study of education and employment found that unem- ployed full-time students take twice as many units as students who are full-time employees and 1.5 times more units than students who are part-time employees.
a. If the fe for the unemployed student is 16 units, what are the fe values for students who work part time and full time?
b. If the student who is unemployed takes 16 units, the student who is em- ployed part time takes 14 units, and the full-time employee takes 12 units, is the expectation supported?
4. In a management trainee program for a multinational corporation, trainees are expected to learn a foreign language. Besides classes at a language training insti- tute, tutors are available. Experience suggests that those learning Japanese seek the help of tutors twice as frequently as those who are learning Spanish. Among 20 students of Japanese, 16 ask for the help of tutors. Among 30 students of Spanish, 8 ask for tutors’ help.
a. What are the fe values? b. Are results consistent with prior experience? c. In this instance, what does HA specify?
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CHAPTER 12Key Terms
5. A marketing analyst is examining the relationship between shoppers’ ethnicity and the purchase of certain grocery item. From ethnic group A, 2 of 12 people purchased the item. From ethnic group B, 5 of 10 people purchased the item. From ethnic group C, 4 of 14 people purchased the item.
a. Are the shoppers’ ethnicities and the tendency to purchase this item inde- pendent?
b. If not, what is the correlation?
6. During the summer months, when electricity usage is high, the power company appeals to customers to reduce consumption by 10% as a public service to avoid blackouts. An alternative is to offer rebates to customers who reduce usage by 10% compared to the same month the previous year. Among 50 randomly selected customers just asked to reduce electricity use, 14 reduce their use by 10% or more. Among 50 randomly selected customers offered rebates, 25 reduce their electricity use by 10%. Are the differences between the public service appeal and the rebates statistically significant?
7. A number of nonprofit groups use fireworks sales as the major fundraiser in the days before the 4th of July. Some of the nonprofits are service groups such as the Veterans of Foreign Wars. Others are intended for support of groups like the cheer- leaders from the local high school. The questions are whether those two groups attract different numbers of customers, and whether the gender of the customer is a factor. Among 20 men who bought fireworks during a particular 2-hour period, 14 purchased from service organizations, the other 6 purchased from non-service groups. Of the 18 women who purchased in the same period, 8 bought from ser- vice organizations, and 10 from non-service groups. Is the gender of the purchaser related to the group from which the purchase is made?
8. A corporate CEO is interested in whether 20 management trainees’ possession of a graduate degree (yes/no) is related to their promotion within the first five years (yes/no). The x2 value is 8.450.
a. What does the x2 value indicate about possession of a graduate degree and promotion?
b. What is the value of f?
Key Terms
• The goodness-of-fit, or 1 3 k, is a chi-square test for significant differences between the frequency observed and the frequency expected in the categories of one nominal- scale variable.
• The chi-square test of independence, or the r 3 k chi-square is a test of whether two nominal scale variables operate independently. The test statistic is the same as for the goodness-of-fit chi-square. A statistically significant result indicates that the two are not independent and a correlation procedure follows.
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CHAPTER 12Key Terms
• The data in a chi-square test of independence are often arranged in a contingency table with the rows indicating the categories of one variable and the columns the categories of the other. With the count of the data entered into the resulting cells, the table provides a visual indicator of the way the variables operate together.
• When a chi-square test of independence is statistically significant, it is followed by a measure of the strength of the correlation between the variables. This is usually a phi coefficient if there are only two categories of one of the variables, or Cramer’s V when there are more than two categories.
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